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How do I show something is pathwise connected?

  • #1
I need to prove that X={(x,y):a<=x<=b, c<=y<=d}

I was thinking of using proof by contradiction.

Assume that X is not pathwise connected, then for a,b in X there is no continuous function that connects the two.

I can show that then the set is disconnected but not sure where to go after that.

Am I going about it the correct way?

Thanks!
 

Answers and Replies

  • #2
22,097
3,279
Why can't you just give the path that connects two points??

Given (x,y) and (u,v). What path can you take that connects (x,y) and (u,v)????
 
  • #3
if I take a straight line path from the points, it would be the distance between the two, so sqrt((x-u)^2 + (y-v)^2)
 
  • #4
22,097
3,279
I don't care about the distance. I want a function [itex][0,1]\rightarrow X[/itex]. A linear path is ok. How do you write that in function language?
 
  • #5
i am not sure what you mean by function language but I would want f(0) = (x,y) and f(1) = (u, v)
 
  • #6
22,097
3,279
Xhat is the equation of the line connecting your two points??
 
  • #7
i want to say something like

given f(0) = (x, y) and p in [0,1] then f(p) = ((y-v)/(x-u)) * p + (x, y)

but thats not correct bc i am switching between R and R^2

do i need to make a continuous function for each component?
 
  • #8
22,097
3,279
Yes, you need to go linearly drom x to u and from y to v.
 
  • #9
do I do it peicewise? like f(p) = (x,y) for p=0, (u, v) for p=1 and then this other functions that I cannot figure out how to get.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,810
934
I need to prove that X={(x,y):a<=x<=b, c<=y<=d}
You need to prove what about this set? That it is connected? Or that it is pathwise connected? A set can be "connected" without being "pathwise connected".

I was thinking of using proof by contradiction.

Assume that X is not pathwise connected, then for a,b in X there is no continuous function that connects the two.

I can show that then the set is disconnected but not sure where to go after that.

Am I going about it the correct way?

Thanks!
 
  • #11
Pathwise connected
 

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